Thrust equation F = ṁv_e + (P_e − P_a)A_e — derivation: Worked examples

3.3.6 · D3 · Physics › Rocket Propulsion › Thrust equation F = ṁv_e + (P_e − P_a)A_e — derivation

Is page mein thrust equation ko $F = \overset{m}{˙} v_{e} + (P_{e} - P_{a}) A_{e}$ har us case se guzara jayega jo exam mein aa sakta hai. Kuch bhi compute karne se pehle, poora map lay out karte hain taaki baad mein koi bhi scenario surprise na kare.

Recall Har symbol ka matlab (from the

parent)

$\overset{m}{˙}$ = exhaust ka mass flow rate, kg/s mein (har second mein kitne kilograms nozzle se nikalta hai).
$v_{e}$ = exhaust speed rocket ke relative, m/s mein.
$P_{e}$ = gas ka static pressure exit plane par, Pa mein.
$P_{a}$ = ambient (bahar ki hawa) pressure, Pa mein.
$A_{e}$ = nozzle exit hole ka area, m² mein.
$F$ = net forward thrust, newtons (N) mein.

Scenario matrix

Is topic ke har problem ka ghar in cells mein se ek hai. "Sign of pressure term" column ko ek kahani ki tarah padho: kya exit gas over-pressured hai, under-pressured hai, ya bahar ke pressure ke saath balanced hai?

#	Cell class	Condition	Sign of $(P_{e} - P_{a}) A_{e}$	Covered by
1	Overexpanded	$P_{e} < P_{a}$	negative (loss)	Ex 1
2	Underexpanded	$P_{e} > P_{a}$	positive (bonus)	Ex 2
3	Perfectly expanded	$P_{e} = P_{a}$	zero	Ex 3
4	Vacuum (degenerate $P_{a} = 0$ )	$P_{a} = 0$	max positive	Ex 4
5	Zero pressure thrust from zero area (degenerate)	$A_{e} \to 0$	zero regardless of $P$	Ex 5
6	Limiting behaviour: thrust vs altitude	$P_{a}$ decreasing	rises as you climb	Ex 6
7	Reverse-solve (find $v_{e}$ or $\overset{m}{˙}$ from $F$ )	given $F$	either	Ex 7
8	Real-world word problem	sea-level launch	negative-ish	Ex 8
9	Exam twist: effective exhaust velocity	uses $v_{e f f}$	folded into one number	Ex 9

Har example neeche us cell(s) ke saath tagged hai jise wo fill karta hai.

Worked examples

Example 1 — Overexpanded nozzle (Cell 1)

Worked example Statement

Sea level par ek first-stage engine: $\overset{m}{˙} = 300$ kg/s, $v_{e} = 2800$ m/s, $P_{e} = 80$ kPa, $P_{a} = 101$ kPa, $A_{e} = 1.2 m^{2}$ . $F$ nikalo.

Forecast: Exit pressure (80 kPa) bahar ki hawa (101 kPa) se kam hai. Guess: pressure thrust negative hoga, isliye $F$ pure momentum thrust se thoda kam niklega. Aage padhne se pehle ek number jot karo.

Step 1 — Momentum thrust. $\overset{m}{˙} v_{e} = 300 \times 2800 = 840, 000 N$ Ye step kyun? Ye Newton's-3rd-law ka core hai: har second mein kitna mass pheka ja raha hai, kitni tez se. Ye hamesha sabse bada term hota hai.

Step 2 — Pressure thrust. $(P_{e} - P_{a}) A_{e} = (80, 000 - 101, 000) (1.2) = (- 21, 000) (1.2) = - 25, 200 N$ Ye step kyun? Kyunki $P_{e} < P_{a}$ hai, bahar ki atmosphere exit region par exhaust se zyada zyar se press karti hai — ek net backward push. Yahi overexpanded penalty hai.

Step 3 — Dono jodo. $F = 840, 000 - 25, 200 = 814, 800 N \approx 814.8 kN$

Verify: Momentum thrust ki units: $(kg/s) (m/s) = kg\cdotm/s^{2} = N$ ✓. Pressure thrust ki units: $(Pa) (m^{2}) = (N/m^{2}) (m^{2}) = N$ ✓. Result forecast ke anusaar 840 kN se thoda neeche hai — atmosphere ne 25.2 kN chura liye.

Example 2 — Underexpanded nozzle (Cell 2)

Worked example Statement

Same $\overset{m}{˙} = 300$ kg/s, $v_{e} = 2800$ m/s, $A_{e} = 1.2 m^{2}$ , lekin zyada altitude par jahan $P_{a} = 30$ kPa hai aur exit gas $P_{e} = 80$ kPa hai. $F$ nikalo.

Forecast: Ab $P_{e} > P_{a}$ hai. Guess: pressure thrust positive ho jayega — ek bonus — isliye $F$ 840 kN se zyada hoga.

Step 1 — Momentum thrust (unchanged, $\overset{m}{˙}$ aur $v_{e}$ same hain): $\overset{m}{˙} v_{e} = 840, 000 N$ Ye step kyun? Engine ne apna throwing rate nahi badla; sirf surrounding air badli hai.

Step 2 — Pressure thrust. $(P_{e} - P_{a}) A_{e} = (80, 000 - 30, 000) (1.2) = (50, 000) (1.2) = 60, 000 N$ Ye step kyun? Exit gas ab bahari thin air se zyada forward push karta hai → net forward force.

Step 3 — Jodo. $F = 840, 000 + 60, 000 = 900, 000 N = 900 kN$

Verify: Same engine, thinner air → zyada thrust, forecast se match karta hai. Example 1 (sea level par 814.8 kN) se compare karo: climb karne se $900 - 814.8 = 85.2$ kN mila. Yahi wajah hai ki rockets upar jaate jaate zyada thrust dete hain.

Example 3 — Perfect expansion (Cell 3)

Worked example Statement

Nozzle is tarah tuned hai ki is altitude par $P_{e} = P_{a} = 50$ kPa exactly hai. $\overset{m}{˙} = 300$ kg/s, $v_{e} = 2800$ m/s, $A_{e} = 1.2 m^{2}$ ke saath, $F$ nikalo.

Forecast: $P_{e} - P_{a} = 0$ , isliye pressure term khatam ho jayega. Guess: $F =$ pure momentum thrust $= 840$ kN.

Step 1 — Pressure thrust. $(P_{e} - P_{a}) A_{e} = (50, 000 - 50, 000) (1.2) = 0 N$ Ye step kyun? Balanced pressures ka matlab hai ki exit region bahar ke saath equilibrium mein hai — koi bhi taraf se koi leftover push nahi.

Step 2 — Total. $F = 840, 000 + 0 = 840, 000 N = 840 kN$

Verify: In chamber conditions ke liye is altitude par ye optimum thrust hai — koi bhi aur $P_{e}$ (over- ya underexpanded) ek alag nozzle mangega aur yahan net useful thrust kam dega. Pressure term literally gayab ho jata hai, isliye $F = \overset{m}{˙} v_{e}$ cleanly milta hai.

Example 4 — Vacuum, the degenerate $P_{a} = 0$ (Cell 4)

Worked example Statement

Engine ko deep space mein le jao: $P_{a} = 0$ . $\overset{m}{˙} = 300$ kg/s, $v_{e} = 2800$ m/s, $P_{e} = 80$ kPa, $A_{e} = 1.2 m^{2}$ rakho. $F$ nikalo.

Forecast: Bahar koi hawa nahi jo push back kare, isliye poori exit pressure ab forward contribute karegi. Guess: ye hamare sab cases mein sabse bada thrust hoga.

Step 1 — Pressure thrust with $P_{a} = 0$ . $(P_{e} - 0) A_{e} = 80, 000 \times 1.2 = 96, 000 N$ Ye step kyun? $P_{a} = 0$ set karne se subtracted back-push bilkul hat jaata hai — "vacuum" ka physical matlab yahi hai. Bahar kuch nahi jo exit gas ka virodh kare.

Step 2 — Total. $F = 840, 000 + 96, 000 = 936, 000 N = 936 kN$

Verify: Ye pichle har case (814.8, 900, 840 kN) se zyada hai ✓. Sea level (Ex 1) se vacuum tak jaate hue, same engine $936 - 814.8 = 121.2$ kN gain karta hai — ye poora $P_{a} A_{e} = 101, 000 \times 1.2 = 121, 200$ N hai jo atmosphere pehle chura leti thi. Isliye linked Specific Impulse ko engine ka "in vacuum" zyada quote kiya jaata hai.

Example 5 — Zero exit area, the degenerate $A_{e} \to 0$ (Cell 5)

Worked example Statement

Ek thought experiment: nozzle exit ko pinhole tak shrink karo, $A_{e} \to 0$ , jabki kisi tarah $\overset{m}{˙} = 300$ kg/s aur $v_{e} = 2800$ m/s maintain rakho. Pressure term ka kya hoga? $P_{e} = 80$ kPa, $P_{a} = 101$ kPa lo.

Forecast: Pressure term mein $A_{e}$ ek factor hai. Guess: pressures chahe kuch bhi hon, agar $A_{e} = 0$ hai toh poora pressure thrust zero hoga.

Step 1 — Pressure thrust. $(P_{e} - P_{a}) A_{e} = (80, 000 - 101, 000) (0) = 0 N$ Ye step kyun? Pressure ek force per area hai; zero area ke saath koi area hi nahi jis par wo act kare. Degenerate case poore second term ko collapse kar deta hai, pressure mismatch chahe kuch bhi ho.

Step 2 — Total. $F = 840, 000 + 0 = 840, 000 N = 840 kN$

Verify: Ek physically real nozzle is flow ko zero area se hold nahi kar sakta, lekin equation sensibly behave karta hai: $A_{e} = 0 \Rightarrow$ sirf momentum thrust bachta hai. Ye confirm karta hai ki pressure thrust purely ek exit-area phenomenon hai, bilkul waise jaise parent derivation ne argue kiya tha (net pressure sirf open exit ke upar survive karta hai).

Example 6 — Limiting behaviour: thrust vs altitude (Cell 6) 📈

Worked example Statement

Engine fix karo ( $\overset{m}{˙} = 300$ kg/s, $v_{e} = 2800$ m/s, $P_{e} = 80$ kPa, $A_{e} = 1.2 m^{2}$ ) aur $P_{a}$ ko rocket climb karne ke saath girne do: $P_{a} = 101, 70, 30, 0$ kPa. $F$ tabulate karo aur trend describe karo.

Forecast: Kam $P_{a}$ → chota subtraction → bada $F$ . Ek smooth, straight-line rise guess karo.

Step 1 — $F$ ko $P_{a}$ ke function ke roop mein likho. $F (P_{a}) = \overset{m}{˙} v_{e} 840, 000 + (80, 000 - P_{a}) (1.2)$ Ye step kyun? Sirf $P_{a}$ badal raha hai, isliye use isolate karo. Baaki sab constant hai. Isse $F$ ek linear function ban jata hai $P_{a}$ ka, slope ke saath $- A_{e} = - 1.2$ N per Pa.

Step 2 — Har altitude par evaluate karo.

$P_{a}$ (kPa)	pressure thrust (N)	$F$ (kN)
101	$(80 - 101) (1.2) \times 1 0^{3} = - 25, 200$	814.8
70	$(80 - 70) (1.2) \times 1 0^{3} = + 12, 000$	852.0
30	$(80 - 30) (1.2) \times 1 0^{3} = + 60, 000$	900.0
0	$(80 - 0) (1.2) \times 1 0^{3} = + 96, 000$	936.0

Ye step kyun? $P_{a}$ ko sea level se vacuum tak sweep karne se pressure term ka har sign ek hi shot mein dikhta hai: 101 par negative, kahi 80 ke paas zero, upar positive.

Figure dekho: red line seedhi hai, dashed "momentum-only" line ko exactly $P_{a} = P_{e} = 80$ kPa par cross karti hai — neeche se upar.

Verify: Crossing point wahan hai jahan pressure thrust $= 0$ hai, yaani $P_{a} = P_{e} = 80$ kPa, jisse $F = 840$ kN milta hai — Example 3 ki logic se match karta hai. Slope check: $P_{a} = 101$ se $P_{a} = 0$ tak, $F$ $101$ kPa ki drop mein $936 - 814.8 = 121.2$ kN bada, yaani $121, 200/101, 000 = 1.2$ N/Pa $= A_{e}$ ✓.

Example 7 — Reverse-solve for exhaust velocity (Cell 7)

Worked example Statement

Ek engine ko sea level par ( $P_{a} = 101$ kPa) $F = 750$ kN deliver karna hai. Uska nozzle $P_{e} = 95$ kPa, $A_{e} = 0.9 m^{2}$ deta hai, aur $\overset{m}{˙} = 260$ kg/s hai. Kaun sa exhaust velocity $v_{e}$ chahiye?

Forecast: Hum equation ko ulta chala rahe hain. $v_{e}$ ke around 3000 m/s guess karo.

Step 1 — Thrust equation ko $v_{e}$ ke liye rearrange karo. $F = \overset{m}{˙} v_{e} + (P_{e} - P_{a}) A_{e} \Rightarrow v_{e} = \frac{F - ( P _{e} - P _{a} ) A _{e}}{m ˙}$ Ye step kyun? $v_{e}$ momentum term ke andar $\overset{m}{˙}$ se multiply hokar baitha hai; (known) pressure term subtract karke aur divide karke use isolate karo.

Step 2 — Pressure term compute karo. $(P_{e} - P_{a}) A_{e} = (95, 000 - 101, 000) (0.9) = (- 6, 000) (0.9) = - 5, 400 N$ Ye step kyun? Thoda overexpanded hai, isliye ye chota negative hai — momentum term ko ye loss puri karni hogi.

Step 3 — Solve karo. $v_{e} = \frac{750 , 000 - ( - 5 , 400 )}{260} = \frac{755 , 400}{260} \approx 2905.4 m/s$ Ye step kyun? Negative subtract karne se add hota hai; engine ko pressure loss cover karne ke liye thoda faster throw karna padega.

Verify: Wapas plug karo: $260 \times 2905.4 = 755, 404$ ; add $- 5, 400 \Rightarrow 750, 004 \approx 750, 000$ N ✓ (rounding). ~3000 m/s ka forecast sahi ballpark mein tha.

Example 8 — Real-world word problem (Cell 8)

Worked example Statement

Ek launch pad engineer sea level par ek test-stand firing se numbers padhti hai: measured thrust $F = 1, 860$ kN, exit pressure $P_{e} = 70$ kPa, ambient $P_{a} = 101$ kPa, exit area $A_{e} = 2.5 m^{2}$ , aur mass flow $\overset{m}{˙} = 620$ kg/s. Wo chahti hai ki effective exhaust velocity jo wo advertise karegi aur kitna thrust atmosphere kha rahi hai.

Forecast: Engine overexpanded hai ( $P_{e} < P_{a}$ ), isliye atmosphere thrust kha rahi hai. Tens of kN ka loss guess karo.

Step 1 — Pressure thrust (atmospheric cost). $(P_{e} - P_{a}) A_{e} = (70, 000 - 101, 000) (2.5) = (- 31, 000) (2.5) = - 77, 500 N$ Ye step kyun? Ye negative number hi atmosphere ka toll hai: pad yahan 77.5 kN thrust khota hai.

Step 2 — Momentum thrust (back it out). $\overset{m}{˙} v_{e} = F - (P_{e} - P_{a}) A_{e} = 1, 860, 000 - (- 77, 500) = 1, 937, 500 N$ Ye step kyun? Total minus pressure se pure Newton's-3rd-law part milta hai.

Step 3 — True relative exhaust velocity. $v_{e} = \frac{1 , 937 , 500}{620} \approx 3125.0 m/s$

Step 4 — Effective exhaust velocity (advertised number). $v_{e f f} = \frac{F}{m ˙} = \frac{1 , 860 , 000}{620} = 3000.0 m/s$ Ye step kyun? $v_{e f f}$ pressure term ko $F = \overset{m}{˙} v_{e f f}$ ke zariye ek figure mein fold karta hai; ye directly Specific Impulse aur Tsiolkovsky Rocket Equation mein jaata hai.

Verify: $v_{e f f} < v_{e}$ (3000 < 3125), jo tab hona hi chahiye jab pressure thrust negative ho ✓. Reconstruct: $\overset{m}{˙} v_{e f f} = 620 \times 3000 = 1, 860, 000 = F$ ✓.

Example 9 — Exam twist: effective exhaust velocity in vacuum (Cell 9)

Worked example Statement

Exam mein $v_{e f f} = 3000$ m/s aur $\overset{m}{˙} = 620$ kg/s sea level par diya gaya hai (Example 8 ke engine se), aur puchha gaya hai: vacuum mein $v_{e f f}$ kya hoga, given $P_{e} = 70$ kPa, $A_{e} = 2.5 m^{2}$ ?

Forecast: Vacuum mein pressure term jump karta hai (koi $P_{a}$ nahi), isliye $v_{e f f}$ barhega. Kuch sau m/s zyada guess karo.

Step 1 — $v_{e f f}$ ki definition yaad karo. $v_{e f f} = v_{e} + \frac{( P _{e} - P _{a} ) A _{e}}{m ˙}$ Ye step kyun? $v_{e f f}$ ka sirf pressure part altitude ke saath badalta hai; $v_{e}$ (relative speed) Example 8 se 3125 m/s par fixed rehta hai.

Step 2 — Vacuum pressure term per unit mass flow. $\frac{( P _{e} - 0 ) A _{e}}{m ˙} = \frac{70 , 000 \times 2.5}{620} = \frac{175 , 000}{620} \approx 282.3 m/s$ Ye step kyun? $P_{a} = 0$ set karne se maximum pressure contribution milta hai, jise $\overset{m}{˙}$ se divide karke velocity units mein convert kiya jaata hai.

Step 3 — $v_{e}$ mein add karo. $v_{e f f, vac} = 3125.0 + 282.3 = 3407.3 m/s$

Verify: Vacuum $v_{e f f}$ (3407.3) $>$ sea-level $v_{e f f}$ (3000) ✓ — engine space mein zyada efficient hai, Example 4 ki story se consistent. Vacuum thrust cross-check: $\overset{m}{˙} v_{e f f, vac} = 620 \times 3407.3 \approx 2, 112, 500$ N; direct: $\overset{m}{˙} v_{e} + P_{e} A_{e} = 1, 937, 500 + 175, 000 = 2, 112, 500$ N ✓.

Recall Quick self-test

Overexpanded kaun sa cell hai aur uske pressure thrust ka sign kya hai? ::: $P_{e} < P_{a}$ , pressure thrust negative Same engine: zyada thrust kahan milega, sea level par ya vacuum mein? ::: Vacuum mein, kyunki $P_{a} = 0$ atmospheric back-push hata deta hai Agar $A_{e} = 0$ ho, toh pressure thrust kya hai? ::: Zero, pressure mismatch chahe kuch bhi ho Measured $F$ se $v_{e}$ kaise nikaaloge? ::: $v_{e} = (F - (P_{e} - P_{a}) A_{e}) / \overset{m}{˙}$ Overexpanded hone par $v_{e f f}$ aur $v_{e}$ ka kya rishta hai? ::: $v_{e f f} < v_{e}$ (pressure term negative hai)

Connections

Parent topic — wo derivation jise ye examples exercise karte hain
Specific Impulse — $v_{e f f}$ se bana (Examples 8, 9)
Tsiolkovsky Rocket Equation — $v_{e f f}$ consume karta hai
De Laval Nozzle — $P_{e}$ aur $A_{e}$ set karta hai (over/under/perfect expansion)
Newton's Third Law — momentum thrust ka origin
Conservation of Momentum — derivation ki foundation